University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.10 - The Binomial Series and Applications of Taylor Series - Practice Exercises - Page 553: 69


$\approx 0.48491743$

Work Step by Step

We know that the Taylor Series for $e^x$ is defined as: $e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$ Now, $e^{(-x^3)}=1+(-x^3)+\dfrac{(-x^3)^2}{2!}+\dfrac{(-x^3)^3}{3!}+\dfrac{(-x^3)^4}{4!}+...=1-x^3+\dfrac{x^6}{2!}-\dfrac{x^9}{3!}+\dfrac{x^{12}}{4!}+... ...(1)$ Take the integral of equation (1). $\int_0^{1/2} e^{(-x^3)}=\int_0^{1/2} [1-x^3+\dfrac{x^6}{2!}-\dfrac{x^9}{3!}+\dfrac{x^{12}}{4!}+..]$ or, $=[x-\dfrac{x^4}{4}+\dfrac{x^7}{(7)2!}-\dfrac{x^{10}}{(10)3!}+\dfrac{x^{13}}{(13)4!}+..]_0^{1/2}$ or, $\dfrac{1}{2}-\dfrac{1}{64}+\dfrac{1}{1792}+....\approx 0.48491743$
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